How do I evaluate $\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$?

Question

$$\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$$

I've tried combining the sum, telescoping series and even trying to make an Nth partial sum but nothing seems to budge. I'm not sure where to go...

Answer 1

Firstly, note that $$\frac{n}{2n-1}-\frac{n+2}{2n+3}=\frac{2n^2+3n-(n+2)(2n-1)}{(2n-1)(2n+3)}$$ which simplifies to $$\frac{2}{(2n-1)(2n+3)}$$ Now use partial fractions to form a telescoping series.

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I hope that helps. If you have any questions please don't hesitate to ask :)

Answer 2

Writing out the first terms explicitly, it is already telescoping, albeit a bit staggered: $$\frac{1}{1}-\frac{3}{5}+\frac{2}{3}-\frac{4}{7}+\frac{3}{5}-\frac{5}{9}+\frac{4}{7}-\frac{6}{11} + ...$$ It looks like $1+\frac{2}{3} = \frac{5}{3}$ is the answer. I suspect that if one of the terms is index shifted by one, the relation would be much more obvious: $$\frac{1}{1}+\frac{2}{3}+\sum_{n=3}^{\infty} \frac{n}{2n-1}-\sum_{n=1}^{\infty}\frac{n+2}{2n+3} = $$ $$\frac{5}{3}+\sum_{k=1}^{\infty}\frac{k+2-2}{2(k+2)+3} -\sum_{n=1}^{\infty}\frac{n+2}{2n+3} $$ EDIT Well, as was pointed out, I was wrong, because the limit terms as the series goes to infinity equal -1. Just goes to show, $\infty - \infty$ can equal anything, so avoid that form when possible. It's really easy to make an answer that looks right, much harder to make it airtight. Kudos to @jjagmath.

Answer 3

We have \begin{align} &\sum _{n=1}^m \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right)=\sum _{n=1}^m \frac{n}{2 n-1}-\sum _{n=1}^m \frac{n+2}{2 (n+2)-1}\\ &=\sum_{n=1}^m \frac{n}{2 n-1}-\sum _{n=3}^{m+2} \frac{n}{2 n-1}\\ &=\frac{1}{2\cdot 1-1}+\frac{2}{2\cdot 2-1}+\sum _{n=3}^m \frac{n}{2 n-1}-\sum _{n=3}^m \frac{n}{2 n-1}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\\ &=1+\frac{2}{3}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\end{align}

Taking the limit when $m \to \infty$ we have \begin{align} &\sum _{n=1}^\infty \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right) \\&= \lim_{m\to\infty}\left( 1+\frac{2}{3}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\right)\\ &= 1+\frac{2}{3}-\frac{1}{2}-\frac{1}{2} = \frac{2}{3}\end{align}

The others answer don't take into account that the "remainder" terms after "telescoping" doesn't tends to $0$.

Answer 4

There is no need of partial fractions since $$\frac{n}{2n-1}=\frac{n-\frac 12+\frac 12}{2n-1}=\frac 12+\frac 1{2(2n-1)}$$ $$\frac{n+2}{2n+3}=\frac{n+\frac 32+\frac 12}{2n+3}=\frac 12+\frac 1{2(2n+3)}$$ $$\frac{n}{2n-1}-\frac{n+2}{2n+3}=\frac 1{2(2n-1)}-\frac 1{2(2n+3)}$$

Answer 5

It becomes easier if you generalise the problem: You are calculating the sum of f(n) - f(n+2), where f(n) = n / (2n-1).

The partial sums for the sum up to k are f(1)-f(3) when k=1, f(1)+f(2)-f(3)-f(4) = f(1)+f(2)-f(k+1)-f(k+2) when k=2, and from then on the partial sum is f(1)+f(2)-f(k+1)-f(k+2) for all k.

In this case, since f(k+1) and f(k+2) both converge towards 1/2, the limit is f(1)+f(2)-1 = 1+2/3-1 = 2/3. The same method will always work if f(k+1)+f(k+2) has a limit, and otherwise the sum doesn’t converge.